Math, asked by Delores5065, 19 days ago

Prove that 7^2n^+1+1 is an integer multiple of 8

Answers

Answered by MysticSohamS
2

Answer:

your solution is as follows

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Step-by-step explanation:

to \: prove \: that :  \\ 7 {}^{2n + 1}  + 1  \: \: is \: a \: multiple \: of \: 8 \\  \\ proof :  \\ we \: would \: prove \: thi s\:mathematical \\  statement \:  \: by \: method \: of \\ mathematical \: induction \\  \\ therefore \: so \: now \\ P \: (n) = 7 {}^{2n + 1}  + 1 \\  \\ Step \:  I :  \\ let \:  \: P(1) = 7 {}^{2(1) + 1}  + 1 \\  \\  = 7 {}^{2 + 1}  + 1 \\  \\  = 7 {}^{3}  + 1 \\  \\  = 343 + 1 \\  \\  = 344 \\  \\  = 8 \times (43) \\  \\ which \: is \: divisible \: and \: multiple \\ by \: 8 \\  \\ hence \: P(1) \: is \: true

Step  \: II :  \\ assume \: that \: P \: (k) \: is \: true \\ ie \: here \:  \\  \\ P(k) = 7 {}^{2k + 1} + 1  = 8a \:  \:  \:  \:  \:  \:  \\  \\  = 7 {}^{2k + 1}  = 8a - 1 \:  \:  \:  \:  \: (1)

Step  \: III :  \\ to \: prove \: that \\ P(k + 1) \:  \: is \: true \\  \\ P(k +1 ) = 7 { }^{2(k + 1) + 1}  + 1  = 8b\\  \\  = 7 {}^{2k + 2 + 1}  + 1 \\  \\  = ( \: 7 {}^{2k + 1} .7 {}^{2}  \: ) + 1 \\  \\  = 49 \times (8a - 1)  + 1 \:  \:  \:  \: from \: (1) \\  \\  =( 49a \times 8) - 49 + 1 \\  \\  = 49a \times 8 - 48 \\  \\  = 8(49a - 6) \\  \\ which \: is \: multiple \: of \: 8 \\ and \: moreove r\: divisible \: by \: it \\  \\ hence \: here \\ P(k + 1) \:  \: is \: true \: for \: all \\ n \: ∈ \: N

hence \: by \:  \\ principle \: of \: mathematical \: induction \\ P(n) \: is \: true

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